Metamath Proof Explorer
		
		
		
		Description:  Cancellation law for subtraction.  (Contributed by NM, 8-Feb-2005)
		
			
				
					|  |  | Ref | Expression | 
					
						|  | Hypotheses | negidi.1 | ⊢ 𝐴  ∈  ℂ | 
					
						|  |  | pncan3i.2 | ⊢ 𝐵  ∈  ℂ | 
					
						|  |  | subadd.3 | ⊢ 𝐶  ∈  ℂ | 
				
					|  | Assertion | subcani | ⊢  ( ( 𝐴  −  𝐵 )  =  ( 𝐴  −  𝐶 )  ↔  𝐵  =  𝐶 ) | 
			
		
		
			
				Proof
				
					
						| Step | Hyp | Ref | Expression | 
						
							| 1 |  | negidi.1 | ⊢ 𝐴  ∈  ℂ | 
						
							| 2 |  | pncan3i.2 | ⊢ 𝐵  ∈  ℂ | 
						
							| 3 |  | subadd.3 | ⊢ 𝐶  ∈  ℂ | 
						
							| 4 |  | subcan | ⊢ ( ( 𝐴  ∈  ℂ  ∧  𝐵  ∈  ℂ  ∧  𝐶  ∈  ℂ )  →  ( ( 𝐴  −  𝐵 )  =  ( 𝐴  −  𝐶 )  ↔  𝐵  =  𝐶 ) ) | 
						
							| 5 | 1 2 3 4 | mp3an | ⊢ ( ( 𝐴  −  𝐵 )  =  ( 𝐴  −  𝐶 )  ↔  𝐵  =  𝐶 ) |